sandydr

Thanks Bluesky, so the error is in the syntax, as I get the following result You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.listid' at line 3 IN select id, title, state, price, fname FROM listings, images JOIN listings.id = images.listid I don't think I am using the correct form to join the tables. Anyone know a good tutorial on this? Regards Sandydr

Thanks for putting me on the right track. I found a tutorial, but still do not quite understand the syntax to use. Modifying the query, it now looks like the below. I however get a 'Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/coralbay/www/www/allsearch/all.php on line 22' - Line 22 being $row=mysql_fetch_array($result); <? $query="select id, title, state, price, fname FROM listings, images join listing.id = images.listid"; $result=mysql_query($query); $cols=4; echo "<table>"; do{ echo "<tr>"; for($i=1;$i<=$cols;$i++){ $row=mysql_fetch_array($result); if($row){ $img = $row['fname']; ?> <td> <table> <tr valign="top"> <td><img src="images/<?=$img ?>" /></td> <td> <b><a href="http://www.coralbayrealestate.com/search/show.php?id=<?=$row['id'] ?> "> <?=$row['title'] ?> </a> </b><br /> <?=$row['state'] ?><br /> $USD <?=number_format ($row['price']) ?><br /> </td> <td width="20"> </td> </tr> </table> </td> <? } else{ echo "<td> </td>"; } } } while($row); echo "</table>"; ?> I have attached a couple of pdf's that show the structure. I tried to copy and paste it in to here, but it was beyond me...! Thanks for you help Sandydr [attachment deleted by admin]

My first post and I am not a programmer or familiar with the correct vocabulary associated with php. I thought I would give the following a go as it is the holidays and my php programmer is no where to be found. I am ok at following scripts to get a simple result, unfortunately the listing manager I am using is coded with Zend. The system is built on a curve2.com and I am pretty much stuck with what exists. I would like to add some more php outputs, which I don't think are too complicated, but I can't seem to get my head around pulling information from multiple tables in to 1 result. I am trying to pull all the data from a 'Listing' table in to 4 columns, which works ok. What I can not get around is trying to link up the 'Image' table to my 'Listing' table and pulling the main image in to my php output. There is a common 'id' but each 'id' has 4+ images and I would like to select the just main one and resize it smaller. So far, I have the following, which I have put together from following examples on this website. 'id' is the unique id for each record. <? $query="select * from listings"; $result=mysql_query($query); $cols=4; echo "<table>"; do{ echo "<tr>"; for($i=1;$i<=$cols;$i++){ $row=mysql_fetch_array($result); if($row){ $img = $row['image_path']; ?> <td> <table> <tr valign="top"> <td><img src="images/<?=$img ?>" /></td> <td> <b><a href="http://www.coralbayrealestate.com/search/show.php?id=<?=$row['id'] ?> "> <?=$row['title'] ?> </a> </b><br /> <?=$row['state'] ?><br /> $USD <?=number_format ($row['price']) ?><br /> </td> <td width="20"> </td> </tr> </table> </td> <? } else{ echo "<td> </td>"; } } } while($row); echo "</table>"; ?> The image table, called 'images' has 3 fields 'id' - individual photos id's 'listid' - Id record to the listing table 'fname' - image name Any guidance would be much appreciated on how to get the primary image on to the table. Happy holidays and best wishes to all for 2011. Sandydr